elemprob-page24 - X i are independent Bernoulli’s 3...

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Proposition 11.2. If X and Y are independent, then Var ( X + Y ) = Var X + Var Y. Proof. We have Var ( X + Y ) = Var X + Var Y + 2Cov ( X, Y ) = Var X + Var Y. Since a binomial is the sum of n independent Bernoulli’s, its variance is np (1 - p ). If we write X = n i =1 X i /n and the X i are independent and have the same distribution ( X is called the sample mean), then E X = E X 1 and Var X = Var X 1 /n . We define the conditional expectation of X given Y by E [ X | Y = y ] = xf X | Y = y ( x ) dx. 12. Moment generating functions. We define the moment generating function m X by m X ( t ) = E e tX , provided this is finite. In the discrete case this is equal to e tx p ( x ) and in the continuous case e tx f ( x ) dx . Let us compute the moment generating function for some of the distributions we have been working with. 1. Bernoulli: pe t + (1 - p ). 2. Binomial: using independence, E e t X i = E e tX i = E e tX i = ( pe t + (1 - p
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Unformatted text preview: X i are independent Bernoulli’s. 3. Poisson: E e tX = X e tk e-λ λ k k ! = e-λ X ( λe t ) k k ! = e-λ e λe t = e λ ( e t-1) . 4. Exponential: E e tX = Z ∞ e tx λe-λx dx = λ λ-t if t < λ and ∞ if t ≥ λ . 5. N (0 , 1): 1 √ 2 π Z e tx e-x 2 / 2 dx = e t 2 / 2 1 √ 2 π Z e-( x-t ) 2 / 2 dx = e t 2 / 2 . 6. N ( μ,σ 2 ): Write X = μ + σZ . Then E e tX = E e tμ e tσZ = e tμ e ( tσ ) 2 / 2 = e tμ + t 2 σ 2 / 2 . Proposition 12.1. If X and Y are independent, then m X + Y ( t ) = m X ( t ) m Y ( t ) . Proof. By independence and Proposition 11.1, m X + Y ( t ) = E e tX e tY = E e tX E e tY = m X ( t ) m Y ( t ) . 24...
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